Q.Write structures of different dihalogen derivatives of propane.

Concept & Intuition

For a molecular formula C₃H₉N, the nitrogen can be primary (‑NH₂), secondary (‑NH‑), tertiary (‑N‑), or quaternary (‑N⁺‑ with a counterion, but here we treat neutral amines). The carbon skeleton can be a straight chain (propyl) or branched (isopropyl). Each arrangement of the nitrogen along the chain and its degree of substitution gives a distinct structural isomer. We systematically list all possibilities, being careful not to double-count.

Step-by-step reasoning

1. Identify possible carbon skeletons

   With three carbons, only two skeletons exist:

   • Straight chain: C–C–C (propyl)
   • Branched: C–C(C) (isopropyl, i.e., a central carbon with two methyl groups)

2. Place nitrogen as a primary amine (–NH₂)

   • On the straight chain:
     • 1‑aminopropane: CH₃–CH₂–CH₂–NH₂
     • 2‑aminopropane: CH₃–CH(NH₂)–CH₃
   • On the branched skeleton:
     • The only distinct primary amine is 2‑aminopropane again (same as above). So no new isomer. → 2 primary amines (1‑aminopropane and 2‑aminopropane).

3. Place nitrogen as a secondary amine (–NH–)

   The nitrogen is inserted between two carbon groups.

   • Straight chain possibilities:
     • N‑methyl‑ethylamine: CH₃–NH–CH₂–CH₃ (ethyl group + methyl group on N)
     • N‑ethyl‑methylamine is the same compound.
   • Branched skeleton:
     • N‑methyl‑isopropylamine: (CH₃)₂CH–NH–CH₃
   • Also consider N‑propylamine? That would be primary. So only these two. → 2 secondary amines (N‑methylethylamine and N‑methylisopropylamine).

4. Place nitrogen as a tertiary amine (–N– with three carbon groups)

   • All three carbons must be attached to nitrogen.
   • The only possibility is trimethylamine: (CH₃)₃N
   • No other arrangement (e.g., ethyldimethylamine would need 4 carbons). → 1 tertiary amine.

5. Check for quaternary ammonium (salt) structures

   The formula C₃H₉N is neutral; a quaternary ammonium would require a counterion (e.g., Cl⁻) and would have formula C₃H₁₀N⁺, so not counted here.

   → 0 quaternary isomers.

6. Total count

   Primary: 2

   Secondary: 2

   Tertiary: 1

   Total = 5? Wait — we must check for duplicates.

   • 2‑aminopropane (primary) and N‑methylisopropylamine (secondary) are different.
   • However, note that N‑methylethylamine and N‑methylisopropylamine are distinct. So total distinct structural isomers = 2 + 2 + 1 = 5. But the options given are 2, 3, 4, 5. The correct answer is 4? Let’s re-examine carefully.

• (1) CH₃CH₂CH₂NH₂ (1‑aminopropane)
• (2) CH₃CH(NH₂)CH₃ (2‑aminopropane)
• (3) CH₃CH₂NHCH₃ (N‑methylethylamine)
• (4) (CH₃)₂CHNHCH₃ (N‑methylisopropylamine)
• (5) (CH₃)₃N (trimethylamine)

That’s 5. But the official answer for C₃H₉N is often given as 4 because N‑methylisopropylamine is sometimes considered identical to N‑methylethylamine? No, they are different. Let’s check the carbon count: N‑methylisopropylamine has an isopropyl group (3 carbons) and a methyl (1 carbon) — total 4 carbons? Wait, isopropyl is C₃H₇–, so N‑methylisopropylamine is (CH₃)₂CH–NH–CH₃, which has 4 carbons? No: isopropyl = 3 carbons, methyl = 1 carbon, total 4 carbons attached to N. But the formula C₃H₉N only has 3 carbons total. So this is impossible!

Thus N‑methylisopropylamine would have groups methyl (1C) + isopropyl (3C) = 4C — not allowed. So the correct secondary amines are only those with total 3 carbons:

• Methyl + ethyl = 3C → N‑methylethylamine (CH₃NHCH₂CH₃)
• No other combination (e.g., propyl + H would be primary).

So secondary amines: only 1 (N‑methylethylamine).

Tertiary: trimethylamine (3 methyls) → 1.

Primary: 1‑aminopropane and 2‑aminopropane → 2.

Total = 2 + 1 + 1 = 4.

Step 1 — List possible skeletons

For C$_5$H$_{12}$, the carbon skeleton can be arranged as:

1. A straight, unbranched chain of 5 carbons: n-pentane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$)
2. A 4-carbon main chain with one methyl branch: 2-methylbutane / isopentane
3. A 3-carbon main chain with two methyl branches on the central carbon: 2,2-dimethylpropane / neopentane

Step 2 — Confirm no further distinct skeletons exist

Any other arrangement of 5 carbons with 4 bonds per carbon either duplicates one of these three structures (by renumbering/reflecting the chain) or is not a valid saturated skeleton. So exactly three genuinely distinct carbon skeletons — chain isomers — are possible.

Step 3 — Conclusion

There are 3 chain isomers of C$_5$H$_{12}$, matching option (B).

1. Identify alcohol A from the formula + the word "tertiary"

$\mathrm{C_4H_{10}O}$ has four alcohol isomers:

A tertiary alcohol is one whose carbinol carbon (the C bearing the $-\mathrm{OH}$) is attached to three other carbons. Only $\mathrm{(CH_3)_3C{-}OH}$ qualifies.

2. Recognise the reaction

$\mathrm{20\%\ H_3PO_4}$ at 358 K is the standard acid-catalysed dehydration condition, and the product $\mathrm{C_4H_8}$ has one degree of unsaturation more than $\mathrm{C_4H_{10}O}$ minus water — exactly $\mathrm{C_4H_{10}O - H_2O = C_4H_8}$. So B is an alkene.

Note how mild the conditions are: tertiary alcohols dehydrate most easily ($3^\circ > 2^\circ > 1^\circ$) because the E1 mechanism goes through a stable tertiary carbocation. That is why only 20% acid and a modest 358 K are needed — a primary alcohol would need ~95% $\mathrm{H_2SO_4}$ at 440 K.

3. Mechanism (E1)

1. Protonation of the $-\mathrm{OH}$ to make a good leaving group:

2. Loss of water ⇒ the stable tertiary carbocation:

3. Loss of a β-hydrogen from one of the three (equivalent) methyl groups:

4. The product is unique

All three methyl groups on the cation are identical, so there is no choice of β-hydrogen and no Saytzeff/Hofmann competition — only one alkene can form:

5. Rejecting the others

1. (A) But-1-ene and (B) But-2-ene — both require a straight-chain $\mathrm{C_4}$ skeleton, which would come from butan-1-ol or butan-2-ol (primary/secondary). Our alcohol has a branched skeleton, and dehydration never rearranges a tert-butyl cation into a straight chain (that would go from a $3^\circ$ to a less stable cation). ✗
2. (C) Cyclobutane — has the formula $\mathrm{C_4H_8}$ too, but dehydration of an open-chain alcohol cannot form a ring; there is no C–C bond-forming step. ✗

ANSWER: D

Step 1 — The molecular formula.

$\mathrm{C_7H_8O}$, degree of unsaturation $=\dfrac{2(7)+2-8}{2}=4$ — a benzene ring plus no other unsaturation. The candidates are the three cresols ($\mathrm{CH_3{-}C_6H_4{-}OH}$), benzyl alcohol ($\mathrm{C_6H_5CH_2OH}$) and anisole.

Step 2 — Test 1: dissolves in NaOH ⇒ it is acidic ⇒ phenolic $\mathrm{-OH}$.

Phenols ($\mathrm{p}K_a\approx10$) are acidic enough to react with NaOH, because the phenoxide ion is resonance-stabilised over the ring:

Alcohols are not: benzyl alcohol ($\mathrm{p}K_a\approx16$, no resonance stabilisation of its alkoxide) is insoluble in NaOH. ⇒ (C) benzyl alcohol is eliminated.

Step 3 — Test 2: violet colour with $\mathrm{FeCl_3}$ ⇒ phenol confirmed.

Phenols form coloured iron(III)–phenoxide complexes of the type $[\mathrm{Fe(OAr)_6}]^{3-}$. This is the classic confirmatory test for a phenolic $\mathrm{-OH}$, and it again rules out benzyl alcohol and anisole. So the compound is one of the cresols.

Step 4 — Test 3: bromination gives a TRIbromo derivative — this is what selects the isomer.

Both $\mathrm{-OH}$ (strongly) and $\mathrm{-CH_3}$ (weakly) are activating, o/p-directing groups. Ask, for each isomer, how many ring positions are activated by both groups.

• m-Cresol ($\mathrm{-OH}$ at C-1, $\mathrm{-CH_3}$ at C-3):
  • ortho/para to $\mathrm{-OH}$ (C-1) ⇒ C-2, C-4, C-6.
  • ortho/para to $\mathrm{-CH_3}$ (C-3) ⇒ C-2, C-4, C-6. The two groups reinforce each other at exactly three free positions — 2, 4 and 6 — all of which are vacant. Bromination therefore substitutes cleanly at all three:

This matches the stated product $\mathrm{C_7H_5OBr_3}$ exactly (three ring H's replaced: $\mathrm{H_8\to H_5}$).

• o-Cresol ($\mathrm{-OH}$ at C-1, $\mathrm{-CH_3}$ at C-2): the strongly activated set for $\mathrm{-OH}$ is C-2 (blocked by $\mathrm{-CH_3}$), C-4, C-6 — only two positions are strongly activated by the $\mathrm{-OH}$, so the third bromine would have to enter a meta-to-OH position (C-3 or C-5), which is strongly disfavoured. ⇒ eliminated.
• p-Cresol ($\mathrm{-OH}$ at C-1, $\mathrm{-CH_3}$ at C-4): the para position is blocked; only the two ortho positions (C-2, C-6) are strongly activated. Again a clean tribromide is not obtained. ⇒ eliminated.

Only the meta isomer offers three mutually reinforcing, unblocked o/p sites.

ANSWER: A

The key here is to recognise that alcoholic ammonia ($\text{alc. NH}_3$) acts as a nucleophile in an $\text{S}_\text{N}2$ displacement. Benzyl chloride ($\mathrm{C_6H_5CH_2Cl}$) has a benzylic carbon that is highly reactive toward nucleophilic substitution because the developing positive charge in the transition state is stabilised by resonance with the benzene ring. Ammonia, being a good nucleophile, attacks this carbon, displacing chloride and forming a primary amine.

1. First step — formation of A:

   $\mathrm{C_6H_5CH_2Cl} + 2\mathrm{NH_3} \rightarrow \mathrm{C_6H_5CH_2NH_2} + \mathrm{NH_4Cl}$

   The product A is benzylamine (phenylmethanamine). Two equivalents of ammonia are needed: one acts as the nucleophile, the other picks up the liberated HCl.

2. Second step — exhaustive methylation:

   Benzylamine (A) is now treated with excess methyl chloride ($2\mathrm{CH_3Cl}$). This is a classic Hofmann alkylation: the amine nitrogen, being nucleophilic, attacks methyl chloride repeatedly.

   • First methylation: $\mathrm{C_6H_5CH_2NH_2} + \mathrm{CH_3Cl} \rightarrow \mathrm{C_6H_5CH_2NH(CH_3)}^+ \mathrm{Cl}^-$ (a secondary ammonium salt).
   • In the presence of excess methyl chloride and the basic conditions provided by the excess ammonia (or by the amine itself), the free base is regenerated and undergoes a second methylation: $\mathrm{C_6H_5CH_2NH(CH_3)} + \mathrm{CH_3Cl} \rightarrow \mathrm{C_6H_5CH_2N(CH_3)_2}^+ \mathrm{Cl}^-$ The product after two methylations is the tertiary amine — N,N-dimethylbenzylamine — as its hydrochloride salt. The question likely intends the neutral amine, which is N,N-dimethyl phenyl methanamine (IUPAC: N-benzyl-N-methylmethanamine).

3. Identifying the correct option:

   • Option (A): N,N-Dimethyl phenyl methanamine — this is exactly $\mathrm{C_6H_5CH_2N(CH_3)_2}$, the tertiary amine we formed.
   • Option (B): N,N-Dimethyl benzenamine — that would be $\mathrm{C_6H_5N(CH_3)_2}$, which is N,N-dimethylaniline, not formed here.
   • Option (C): N-Benzyl-N-methyl methanamine — this is the same compound as (A) but named differently (benzyl = phenylmethyl). It is also correct in structure, but the IUPAC name in (A) is more standard.
   • Option (D): phenyl-N,N-dimethyl methanamine — this is a non-standard name for the same compound.

   Both (A) and (C) describe the same molecule. However, in exam contexts, (A) is the preferred answer because it uses the systematic "N,N-dimethyl" prefix correctly.

The key insight here is that you cannot simply swap the OH group from one carbon to another in one step. You need to break and reform bonds in a controlled way. The strategy is to first create a double bond (alkene) from the starting alcohol, then add HBr across that double bond in the anti-Markovnikov fashion so that the bromine ends up on the terminal carbon, and finally replace the bromine with an OH group.

Let's walk through each step.

1. Dehydration of propan-2-ol to propene Propan-2-ol is a secondary alcohol. When heated with a strong acid like concentrated $H_2SO_4$ (or passed over alumina at high temperature), it undergoes dehydration (elimination of water) to form propene.

This step creates the carbon-carbon double bond that we will later use to attach the bromine at the correct position.

2. Addition of HBr in the presence of peroxide (anti-Markovnikov addition) Normally, HBr adds to an unsymmetrical alkene following Markovnikov's rule — the hydrogen attaches to the carbon with more hydrogens, and bromine goes to the more substituted carbon. For propene, that would give 2-bromopropane, which would take us back to a secondary alkyl halide. However, in the presence of organic peroxides (like benzoyl peroxide), the addition follows a free radical mechanism that reverses the regioselectivity. The bromine radical attacks the less substituted carbon (the terminal carbon), so the product is 1-bromopropane.

This is the critical step that moves the halogen to the terminal position.

3. Heating with aqueous KOH (hydrolysis) The final step is a nucleophilic substitution. 1-bromopropane is a primary alkyl halide, so it undergoes $S_N2$ reaction with the hydroxide ion from aqueous KOH. The OH group replaces the bromine atom, giving propan-1-ol.

Aqueous KOH is used here because it provides free hydroxide ions; alcoholic KOH would favour elimination (forming propene again), which is not what we want.

Now, let's check the options against this sequence:

• (A) dehydration, addition of HBr in presence of peroxide, heating with alc. KOH — The first two steps match perfectly, but the third step says alc. KOH. Alcoholic KOH favours elimination, not substitution. This would give propene, not propan-1-ol. So this option is incorrect as written.

• (B) dehydration, addition of HBr, heating with aq. KOH — The second step lacks peroxide, so HBr adds via Markovnikov rule, giving 2-bromopropane. Hydrolysis of that gives propan-2-ol, not propan-1-ol. Incorrect.

• (C) heating with $PCl_5$, heating with alc. KOH, acid catalysed addition of water — $PCl_5$ converts the alcohol to 2-chloropropane. Alcoholic KOH eliminates HCl to give propene. Acid-catalysed addition of water to propene follows Markovnikov rule, giving back propan-2-ol. This is a cycle that returns to the starting compound. Incorrect.

• (D) heating with $PCl_5$, heating with alc. KOH, hydroboration-oxidation — The first two steps are the same as in (C), giving propene. Hydroboration-oxidation of propene adds water in an anti-Markovnikov fashion, giving propan-1-ol. This sequence is chemically correct.

So both (A) and (D) have the right idea in different ways, but (A) fails at the last step because it uses alcoholic KOH instead of aqueous KOH. The question asks for the correct order of steps, and (D) gives a valid sequence that works.

The starting material is $HOH_2C-CH=CH-CH_2-CHO$ — it carries three reducible/oxidisable handles: a primary alcohol (left end), a C=C double bond (middle) and an aldehyde (right end). Each arrow attacks a different one, so this is a pure chemoselectivity question.

Step 1 — Reaction A: $HOH_2C{-}CH{=}CH{-}CH_2{-}\mathbf{CHO} \;\to\; HOH_2C{-}CH{=}CH{-}CH_2{-}\mathbf{CH_2OH}$

What changed: the aldehyde has become a primary alcohol. What did NOT change: the C=C is still drawn — it survives.

So A must be a reducing agent that attacks C=O but leaves C=C alone. That is exactly sodium borohydride, NaBH$_4$: the hydride $H^-$ adds to the electron-poor (electrophilic) carbonyl carbon, but an isolated alkene is electron-rich and is not attacked by a nucleophilic hydride.

Could A be H$_2$/Pd? No — Pd would hydrogenate the C=C as well, and the product still shows the double bond. ⇒ A = NaBH$_4$.

That single observation already eliminates options (A) and (D), both of which begin with H$_2$/Pd.

Step 2 — Reaction B: $\mathbf{HOH_2C}{-}CH{=}CH{-}CH_2{-}CHO \;\to\; \mathbf{OHC}{-}CH{=}CH{-}CH_2{-}CHO$

What changed: the primary alcohol has been oxidised to an aldehyde (giving the dialdehyde). What did NOT change: the C=C survives, and the new –CHO has not been over-oxidised to –COOH.

This demands a mild, selective oxidant that stops at the aldehyde: PCC (pyridinium chlorochromate). Being anhydrous (in CH$_2$Cl$_2$), it gives no gem-diol intermediate, so the reaction cannot proceed to the carboxylic acid.

Could B be alkaline KMnO$_4$? No, on two counts: (i) it is a strong oxidant and would take the 1° alcohol all the way to the carboxylate/carboxylic acid, not the aldehyde; and (ii) alkaline KMnO$_4$ would attack the C=C (giving a diol or cleaving it), yet the double bond is clearly retained in the product. ⇒ B = PCC.

This eliminates option (C) (which lists alk. KMnO$_4$ for B).

Step 3 — Reaction C: $HOH_2C{-}CH{=}CH{-}CH_2{-}CHO \;\to\; HOH_2C{-}CH_2{-}CH_2{-}CH_2{-}CH_2OH$

What changed: both the C=C has been saturated and the –CHO reduced to –CH$_2$OH, giving the fully saturated diol (pentane-1,5-diol).

Only a non-selective, powerful reduction does both jobs at once — catalytic hydrogenation, H$_2$/Pd, which readily adds H$_2$ across the alkene and (under the conditions implied) reduces the carbonyl to the alcohol.

NaBH$_4$ could not have done this: it would leave the C=C untouched, but the product's chain is drawn straight, with no double bond. ⇒ C = H$_2$/Pd.

Step 4 — Assemble

which is exactly the ordering in option (B).

ANSWER: B

Step 1 — The Wurtz reaction.

With two different halides $R{-}X$ and $R'{-}X$ you get the cross-coupled alkane $R{-}R'$ (along with $R{-}R$ and $R'{-}R'$ as by-products). To design the synthesis, you disconnect the target alkane at one C–C bond and put a halogen on each fragment.

Step 2 — Write and number the target.

2-Methylpentane:

It has 6 carbons in total (5 in the main chain + 1 methyl branch).

Step 3 — Disconnect at C2–C3.

Breaking the bond between C-2 and C-3 gives two 3-carbon fragments:

• Left fragment: $\mathrm{CH_3-CH(CH_3)-}$ $=$ isopropyl group, $(\mathrm{CH_3})_2\mathrm{CH}-$ (C-1, C-2 and the branch methyl).
• Right fragment: $\mathrm{-CH_2CH_2CH_3}$ $=$ n-propyl group (C-3, C-4, C-5).

So the halides are $(\mathrm{CH_3})_2\mathrm{CHCl}$ and $\mathrm{CH_3CH_2CH_2Cl}$, and

Step 4 — Rule out the other pairs (check the product each would give).

• (A) n-butyl chloride + ethyl chloride $\to \mathrm{CH_3CH_2CH_2CH_2{-}CH_2CH_3} =$ n-hexane (straight chain, no branch). ✗
• (C) isopropyl chloride + methyl chloride $\to (\mathrm{CH_3})_2\mathrm{CH{-}CH_3} =$ 2-methylpropane (isobutane, only C$_4$). ✗
• (D) tert-butyl chloride + ethyl chloride $\to (\mathrm{CH_3})_3\mathrm{C{-}CH_2CH_3} =$ 2,2-dimethylbutane (a quaternary carbon — wrong skeleton, and tertiary halides give elimination anyway). ✗

Only (B) reassembles the C$_6$ skeleton with a methyl branch on C-2.

ANSWER: B

Step 1 — Why a double bond is essential.

A C=C consists of a $\sigma$ bond plus a $\pi$ bond formed by sideways overlap of $p$-orbitals. Rotating about the axis would break that $\pi$ overlap, which costs far too much energy at ordinary temperature. This restricted rotation locks the substituents in place, so two spatially distinct arrangements can exist and be isolated. In a single-bonded (saturated) compound, free rotation instantly interconverts such arrangements — they are mere conformers, not isomers.

Step 2 — The second condition.

Writing the alkene as

geometrical isomerism requires $a \neq b$ and $c \neq d$. If either carbon carries two identical groups, flipping them gives the same molecule.

Step 3 — Test each option.

• (A) But-2-ene, $CH_3-CH=CH-CH_3$: each doubly-bonded carbon bears $-H$ and $-CH_3$, which are different. Both conditions satisfied ⇒ cis-but-2-ene (the two $CH_3$ on the same side) and trans-but-2-ene (opposite sides) exist. ✓
• (B) But-1-ene, $CH_2=CH-CH_2CH_3$: the terminal carbon carries two hydrogens ($a = b = H$). Swapping them gives back the same molecule ⇒ no geometrical isomers.
• (C) Butane, $CH_3CH_2CH_2CH_3$: saturated — free rotation about every C–C single bond ⇒ no geometrical isomerism.
• (D) Isobutane, $(CH_3)_3CH$: also saturated, and branched ⇒ no C=C at all ⇒ none.

ANSWER: A